How to evaluate $lim_{xto 0} frac {(sin(2x)-2sin(x))^4}{(3+cos(2x)-4cos(x))^3}$?
$begingroup$
$$lim_{xto 0} frac {(sin(2x)-2sin(x))^4}{(3+cos(2x)-4cos(x))^3}$$
without L'Hôpital.
I've tried using equivalences with ${(sin(2x)-2sin(x))^4}$ and arrived at $-x^{12}$ but I don't know how to handle ${(3+cos(2x)-4cos(x))^3}$. Using $cos(2x)=cos^2(x)-sin^2(x)$ hasn't helped, so any hint?
real-analysis limits-without-lhopital
edited Jan 8 at 18:17
Andrei
11.6k21026
asked Jan 8 at 18:14
iggykimiiggykimi
1179
$endgroup$
add a comment |
$begingroup$
$$lim_{xto 0} frac {(sin(2x)-2sin(x))^4}{(3+cos(2x)-4cos(x))^3}$$
without L'Hôpital.
I've tried using equivalences with ${(sin(2x)-2sin(x))^4}$ and arrived at $-x^{12}$ but I don't know how to handle ${(3+cos(2x)-4cos(x))^3}$. Using $cos(2x)=cos^2(x)-sin^2(x)$ hasn't helped, so any hint?
real-analysis limits-without-lhopital
edited Jan 8 at 18:17
Andrei
11.6k21026
asked Jan 8 at 18:14
iggykimiiggykimi
1179
$endgroup$
add a comment |
$begingroup$
$$lim_{xto 0} frac {(sin(2x)-2sin(x))^4}{(3+cos(2x)-4cos(x))^3}$$
without L'Hôpital.
I've tried using equivalences with ${(sin(2x)-2sin(x))^4}$ and arrived at $-x^{12}$ but I don't know how to handle ${(3+cos(2x)-4cos(x))^3}$. Using $cos(2x)=cos^2(x)-sin^2(x)$ hasn't helped, so any hint?
real-analysis limits-without-lhopital
edited Jan 8 at 18:17
Andrei
11.6k21026
asked Jan 8 at 18:14
iggykimiiggykimi
1179
$endgroup$
$$lim_{xto 0} frac {(sin(2x)-2sin(x))^4}{(3+cos(2x)-4cos(x))^3}$$
without L'Hôpital.
I've tried using equivalences with ${(sin(2x)-2sin(x))^4}$ and arrived at $-x^{12}$ but I don't know how to handle ${(3+cos(2x)-4cos(x))^3}$. Using $cos(2x)=cos^2(x)-sin^2(x)$ hasn't helped, so any hint?
real-analysis limits-without-lhopital
real-analysis limits-without-lhopital
edited Jan 8 at 18:17
Andrei
11.6k21026
asked Jan 8 at 18:14
iggykimiiggykimi
1179
edited Jan 8 at 18:17
Andrei
11.6k21026
asked Jan 8 at 18:14
iggykimiiggykimi
1179
edited Jan 8 at 18:17
Andrei
11.6k21026
edited Jan 8 at 18:17
Andrei
11.6k21026
edited Jan 8 at 18:17
Andrei
11.6k21026
11.6k21026
asked Jan 8 at 18:14
iggykimiiggykimi
1179
asked Jan 8 at 18:14
iggykimiiggykimi
1179
asked Jan 8 at 18:14
iggykimiiggykimi
1179
1179
add a comment |
add a comment |
3 Answers
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$begingroup$
Hint: Note that
$$ 3+cos(2x)-4cos(x) = 3 + 2cos^2(x) - 1 - 4cos(x) = 2(cos(x)-1)^2, $$
and that
$$ sin(2x) - 2sin(x) = 2sin(x)cos(x)-2sin(x) = 2sin(x)(cos(x)-1). $$
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8451624
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$begingroup$
Hint: Your quotient can be simplified to $$8cosleft(frac{x}{2}right)^4$$
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
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$begingroup$
One can evaluate all such lims using series expansions.
$$
sin(x) = x - frac{x^3}{6} + o(x^4)\
cos(x) = 1 - frac{x^2}{2} + o(x^4)
$$
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.
L'Hôpital's rule is another form of this more general approach.
answered Jan 8 at 18:30
Mike_Mike_
366
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Note that
$$ 3+cos(2x)-4cos(x) = 3 + 2cos^2(x) - 1 - 4cos(x) = 2(cos(x)-1)^2, $$
and that
$$ sin(2x) - 2sin(x) = 2sin(x)cos(x)-2sin(x) = 2sin(x)(cos(x)-1). $$
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
$begingroup$
Hint: Note that
$$ 3+cos(2x)-4cos(x) = 3 + 2cos^2(x) - 1 - 4cos(x) = 2(cos(x)-1)^2, $$
and that
$$ sin(2x) - 2sin(x) = 2sin(x)cos(x)-2sin(x) = 2sin(x)(cos(x)-1). $$
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
$begingroup$
Hint: Note that
$$ 3+cos(2x)-4cos(x) = 3 + 2cos^2(x) - 1 - 4cos(x) = 2(cos(x)-1)^2, $$
and that
$$ sin(2x) - 2sin(x) = 2sin(x)cos(x)-2sin(x) = 2sin(x)(cos(x)-1). $$
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8451624
$endgroup$
Hint: Note that
$$ 3+cos(2x)-4cos(x) = 3 + 2cos^2(x) - 1 - 4cos(x) = 2(cos(x)-1)^2, $$
and that
$$ sin(2x) - 2sin(x) = 2sin(x)cos(x)-2sin(x) = 2sin(x)(cos(x)-1). $$
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8451624
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8451624
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8451624
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8451624
5,8451624
add a comment |
add a comment |
$begingroup$
Hint: Your quotient can be simplified to $$8cosleft(frac{x}{2}right)^4$$
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
$endgroup$
add a comment |
$begingroup$
Hint: Your quotient can be simplified to $$8cosleft(frac{x}{2}right)^4$$
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
$endgroup$
add a comment |
$begingroup$
Hint: Your quotient can be simplified to $$8cosleft(frac{x}{2}right)^4$$
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
$endgroup$
Hint: Your quotient can be simplified to $$8cosleft(frac{x}{2}right)^4$$
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
73.7k42864
add a comment |
add a comment |
$begingroup$
One can evaluate all such lims using series expansions.
$$
sin(x) = x - frac{x^3}{6} + o(x^4)\
cos(x) = 1 - frac{x^2}{2} + o(x^4)
$$
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.
L'Hôpital's rule is another form of this more general approach.
answered Jan 8 at 18:30
Mike_Mike_
366
$endgroup$
add a comment |
$begingroup$
One can evaluate all such lims using series expansions.
$$
sin(x) = x - frac{x^3}{6} + o(x^4)\
cos(x) = 1 - frac{x^2}{2} + o(x^4)
$$
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.
L'Hôpital's rule is another form of this more general approach.
answered Jan 8 at 18:30
Mike_Mike_
366
$endgroup$
add a comment |
$begingroup$
One can evaluate all such lims using series expansions.
$$
sin(x) = x - frac{x^3}{6} + o(x^4)\
cos(x) = 1 - frac{x^2}{2} + o(x^4)
$$
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.
L'Hôpital's rule is another form of this more general approach.
answered Jan 8 at 18:30
Mike_Mike_
366
$endgroup$
One can evaluate all such lims using series expansions.
$$
sin(x) = x - frac{x^3}{6} + o(x^4)\
cos(x) = 1 - frac{x^2}{2} + o(x^4)
$$
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.
L'Hôpital's rule is another form of this more general approach.
answered Jan 8 at 18:30
Mike_Mike_
366
answered Jan 8 at 18:30
Mike_Mike_
366
answered Jan 8 at 18:30
Mike_Mike_
366
answered Jan 8 at 18:30
Mike_Mike_
366
366
add a comment |
add a comment |
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